\(b,\sqrt{36}.\sqrt{\dfrac{25}{26}}+\dfrac{1}{4}\\ =\sqrt{6^2}.\sqrt{\left(\dfrac{5}{4}\right)^2}+\dfrac{1}{4}\\=6.\dfrac{5}{4}+\dfrac{1}{4}=\dfrac{30}{4}+\dfrac{1}{4}=\dfrac{31}{4}\)
\(c,\sqrt{\dfrac{4}{81}}:\sqrt{\dfrac{25}{81}}-1\dfrac{2}{5}\\ =\sqrt{\left(\dfrac{2}{9}\right)^2}:\sqrt{\left(\dfrac{5}{9}\right)^2}-\dfrac{7}{5}\\ =\dfrac{2}{9}:\dfrac{5}{9}-\dfrac{7}{5}\\ =\dfrac{2}{9}.\dfrac{9}{5}-\dfrac{7}{5}=\dfrac{2}{5}-\dfrac{7}{5}\\ =-1\)
\(d, 0,1.\sqrt{225}.\sqrt{\dfrac{1}{4}}\\ =\dfrac{1}{10}.\sqrt{15^2}.\sqrt{\left(\dfrac{1}{2}\right)^2}\\ =\dfrac{1}{10}.15.\dfrac{1}{2}=\dfrac{3}{5}\)
\(e, \dfrac{3^{25}}{9^3.3^{16}}\\ =\dfrac{3^{25}}{\left(3^2\right)^3.3^{16}}\\ =\dfrac{3^{25}}{3^6.3^{16}}\\ =\dfrac{3^{25}}{3^{22}}\\ =3^3=27\)
Giúp mình với,mình cần gấp!Cảm ơn trc ạ.
